glob的排序根据

qxt912

请教：glob是通过哪种方式，access,modify,还是change时间戳对得到的文件排序的？

jason680

回复 1# qxt912

$ perldoc -f glob
    glob EXPR
    glob    In list context, returns a (possibly empty) list of filename
            expansions on the value of EXPR such as the standard Unix shell
            /bin/csh would do. In scalar context, glob iterates through such
            filename expansions, returning undef when the list is exhausted.
            This is the internal function implementing the "<*.c>" operator,
            but you can use it directly. If EXPR is omitted, $_ is used. The
            "<*.c>" operator is discussed in more detail in "I/O Operators" in
            perlop.

            Note that "glob" splits its arguments on whitespace and treats
            each segment as separate pattern. As such, "glob("*.c *.h")"
            matches all files with a .c or .h extension. The expression
            "glob(".* *")" matches all files in the current working directory.

            If non-empty braces are the only wildcard characters used in the
            "glob", no filenames are matched, but potentially many strings are
            returned. For example, this produces nine strings, one for each
            pairing of fruits and colors:

                @many =  glob "{apple,tomato,cherry}={green,yellow,red}";

            Beginning with v5.6.0, this operator is implemented using the
            standard "File::Glob" extension. See File::Glob for details,
            including "bsd_glob" which does not treat whitespace as a pattern
            separator.
   

qxt912

回复 2# jason680


    具体那句话写出了glob 排序的根据。。

jason680

回复 3# qxt912

glob without sort
   

rongchaogao

完全可以读入到数组以后，按照你自己想要的方式排序。